The equation obtained by transforming $x^2+y^2-6x+10y-2=0$ to the parallel axes through $(3,-5)$ is

$A$ circle is drawn with the $y$-axis as a tangent and its center at the point which is the reflection of $(3, 4)$ in the line $y = x$. The equation of the circle is:

The centre of the circle given by the parametric equations $x = -1 + 2\cos \theta$ and $y = 3 + 2\sin \theta$ is:

The equation of the circumcircle of the quadrilateral formed by the lines $x = a, x = 2a, y = -a, y = a$ is:

The lines $2x - 3y = 5$ and $3x - 4y = 7$ are the diameters of a circle of area $154$ square units. The equation of the circle is

The parametric equations of the circle $x^2+y^2-6x-2y+9=0$ are